Understanding the Problem

The question asks for the value of the sum of the cosines of the angles of a triangle ABC, given its side lengths. The side lengths are AB = 16 m, BC = 10 m, and CA = 10 m. We need to find the value of \( \cos A + \cos B + \cos C \).

We are also given information about a lamp post situated at the middle point of side AB and the angle it subtends at vertex B. However, the question specifically asks for the sum of the cosines of the triangle's angles, which can be determined directly from the side lengths using the Law of Cosines. The lamp post information appears to be additional data not required for this specific calculation.

Analyzing the Triangle ABC

The triangle ABC has side lengths:

• a (opposite vertex A) = BC = 10 m
• b (opposite vertex B) = CA = 10 m
• c (opposite vertex C) = AB = 16 m

Since BC = CA (a = b), the triangle ABC is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, angle A is equal to angle B.

Using the Law of Cosines to Find Angle Cosines

The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formulas are:

• \(a^2 = b^2 + c^2 - 2bc \cos A\)
• \(b^2 = a^2 + c^2 - 2ac \cos B\)
• \(c^2 = a^2 + b^2 - 2ab \cos C\)

We can rearrange these formulas to find the cosine of each angle:

• \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
• \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}\)
• \(\cos C = \frac{a^2 + b^2 - c^2}{2ab}\)

Calculating cos A

Using the formula for \(\cos A\):

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substitute the side lengths a=10, b=10, c=16:

\[ \cos A = \frac{10^2 + 16^2 - 10^2}{2 \times 10 \times 16} \]

\[ \cos A = \frac{100 + 256 - 100}{320} \]

\[ \cos A = \frac{256}{320} \]

Simplify the fraction:

\[ \cos A = \frac{256 \div 64}{320 \div 64} = \frac{4}{5} \]

So, \( \cos A = \frac{4}{5} \).

Calculating cos B

Since triangle ABC is isosceles with a=b, angle A = angle B. Therefore, \( \cos B = \cos A \).

\[ \cos B = \frac{4}{5} \]

Alternatively, using the formula for \(\cos B\):

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

Substitute the side lengths a=10, b=10, c=16:

\[ \cos B = \frac{10^2 + 16^2 - 10^2}{2 \times 10 \times 16} \]

\[ \cos B = \frac{100 + 256 - 100}{320} \]

\[ \cos B = \frac{256}{320} = \frac{4}{5} \]

So, \( \cos B = \frac{4}{5} \).

Calculating cos C

Using the formula for \(\cos C\):

\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

Substitute the side lengths a=10, b=10, c=16:

\[ \cos C = \frac{10^2 + 10^2 - 16^2}{2 \times 10 \times 10} \]

\[ \cos C = \frac{100 + 100 - 256}{200} \]

\[ \cos C = \frac{200 - 256}{200} \]

\[ \cos C = \frac{-56}{200} \]

Simplify the fraction:

\[ \cos C = \frac{-56 \div 8}{200 \div 8} = -\frac{7}{25} \]

So, \( \cos C = -\frac{7}{25} \).

Calculating cos A + cos B + cos C

Now we need to sum the values we found for \( \cos A \), \( \cos B \), and \( \cos C \):

\[ \cos A + \cos B + \cos C = \frac{4}{5} + \frac{4}{5} + \left(-\frac{7}{25}\right) \]

\[ \cos A + \cos B + \cos C = \frac{8}{5} - \frac{7}{25} \]

To subtract the fractions, find a common denominator, which is 25. Convert \( \frac{8}{5} \) to an equivalent fraction with denominator 25:

\[ \frac{8}{5} = \frac{8 \times 5}{5 \times 5} = \frac{40}{25} \]

Now perform the subtraction:

\[ \frac{40}{25} - \frac{7}{25} = \frac{40 - 7}{25} = \frac{33}{25} \]

Therefore, \( \cos A + \cos B + \cos C = \frac{33}{25} \).

Relevance of the Lamp Post Information

The information about the lamp post being at the midpoint of AB and subtending an angle of 45° at vertex B is not necessary for calculating the sum of the cosines of the triangle's angles. The angles of a triangle are determined solely by its side lengths. This extra information might be used in a different part of a larger problem (e.g., finding the height of the lamp post), but it does not affect the internal angles of triangle ABC itself.

Conclusion

By applying the Law of Cosines to the given side lengths of triangle ABC, we calculated \( \cos A \), \( \cos B \), and \( \cos C \). Summing these values gives the result.

Revision Table: Triangle Angle Calculations

Additional Information: Properties of Triangles

• The sum of the angles in any triangle is always 180° (\( A + B + C = 180^\circ \)).
• For any triangle, there is a relationship between the sum of the cosines of its angles: \( \cos A + \cos B + \cos C = 1 + 4 \sin(A/2) \sin(B/2) \sin(C/2) \). While this identity exists, calculating the individual cosines using the Law of Cosines and summing them is a direct method when side lengths are known.
• The Law of Sines is another fundamental law relating side lengths and angles: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
• In an isosceles triangle, the altitude from the vertex angle bisects the base and also bisects the vertex angle.